sin^2(ax^2)-sin^2(bx^2)[0,∞]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\infty} \sin ax^2 \cos bx^2 dx=
\begin{cases}
\displaystyle \frac{1}{4}\sqrt{\frac{π}{2}}\left(\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{a-b}}\right) (a \gt b)\\
\displaystyle \frac{1}{4}\sqrt{\frac{π}{2}}\left(\frac{1}{\sqrt{b+a}}-\frac{1}{\sqrt{b-a}}\right) (b \gt a)\\
\end{cases}\\
&(2) \displaystyle\int_0^{\infty} (\sin^2 ax^2 -\sin^2 bx^2)dx=\frac{1}{8}\left(\sqrt{\frac{π}{b}}-\sqrt{\frac{π}{a}}\right)\\
&(3) \displaystyle\int_0^{\infty} (\cos^2 ax^2 -\sin^2 bx^2)dx=\frac{1}{8}\left(\sqrt{\frac{π}{b}}+\sqrt{\frac{π}{a}}\right)\\
&(4) \displaystyle\int_0^{\infty} (\cos^2 ax^2 -\cos^2 bx^2)dx=\frac{1}{8}\left(\sqrt{\frac{π}{a}}-\sqrt{\frac{π}{b}}\right)\\
\end{alignat}
ただし \(a \gt 0, b \gt 0\)





<証明>

\((1)\) から \((4)\) まで半角の公式(または積和の公式)で次数を下げて、
次のフレネル積分を用います。$$\displaystyle\int_0^{\infty} \cos ax^2 dx=\displaystyle\int_0^{\infty} \sin ax^2 dx=\frac{1}{2}\sqrt{\frac{π}{2a}}$$

$$(1) \displaystyle\int_0^{\infty} \sin ax^2 \cos bx^2 dx=\frac{1}{2}\displaystyle\int_0^{\infty} \{\sin (a+b)x^2+ \sin (a-b)x^2\}dx$$
\((A)\) \(a \gt b\) のとき
\begin{alignat}{2}
&=\frac{1}{2}\left\{\frac{1}{2}\sqrt{\frac{π}{2(a+b)}}+\frac{1}{2}\sqrt{\frac{π}{2(a-b)}}\right\}\\
&=\frac{1}{4}\sqrt{\frac{π}{2}}\left(\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{a-b}}\right)
\end{alignat}
\((B)\) \(b \gt a\) のとき
\begin{alignat}{2}
&=\frac{1}{2}\displaystyle\int_0^{\infty} \{\sin (b+a)x^2- \sin (b-a)x^2\}dx\\
&=\frac{1}{2}\left\{\frac{1}{2}\sqrt{\frac{π}{2(b+a)}}-\frac{1}{2}\sqrt{\frac{π}{2(b-a)}}\right\}\\
&=\frac{1}{4}\sqrt{\frac{π}{2}}\left(\frac{1}{\sqrt{b+a}}-\frac{1}{\sqrt{b-a}}\right)
\end{alignat}







\begin{alignat}{2}
&(2) \displaystyle\int_0^{\infty} (\sin^2 ax^2 -\sin^2 bx^2)dx=\frac{1}{2}\displaystyle\int_0^{\infty} \{1- \cos 2ax^2-(1-\cos 2bx^2)\}dx\\
&                          =\frac{1}{2}\displaystyle\int_0^{\infty} (\cos 2bx^2-\cos 2ax^2)dx\\
&                          =\frac{1}{2}\left(\frac{1}{2}\sqrt{\frac{π}{4b}}-\frac{1}{2}\sqrt{\frac{π}{4a}}\right)\\
&                          =\frac{1}{8}\left(\sqrt{\frac{π}{b}}-\sqrt{\frac{π}{a}}\right)
\end{alignat}







\begin{alignat}{2}
&(3) \displaystyle\int_0^{\infty} (\cos^2 ax^2 -\sin^2 bx^2)dx=\frac{1}{2}\displaystyle\int_0^{\infty} \{1+ \cos 2ax^2-(1-\cos 2bx^2)\}dx\\
&                          =\frac{1}{2}\displaystyle\int_0^{\infty} (\cos 2bx^2+\cos 2ax^2)dx\\
&                          =\frac{1}{2}\left(\frac{1}{2}\sqrt{\frac{π}{4b}}+\frac{1}{2}\sqrt{\frac{π}{4a}}\right)\\
&                          =\frac{1}{8}\left(\sqrt{\frac{π}{b}}+\sqrt{\frac{π}{a}}\right)
\end{alignat}







\begin{alignat}{2}
&(4) \displaystyle\int_0^{\infty} (\cos^2 ax^2 -\cos^2 bx^2)dx=\frac{1}{2}\displaystyle\int_0^{\infty} \{1+ \cos 2ax^2-(1+\cos 2bx^2)\}dx\\
&                          =\frac{1}{2}\displaystyle\int_0^{\infty} (\cos 2ax^2-\cos 2bx^2)dx\\
&                          =\frac{1}{2}\left(\frac{1}{2}\sqrt{\frac{π}{4a}}-\frac{1}{2}\sqrt{\frac{π}{4b}}\right)\\
&                          =\frac{1}{8}\left(\sqrt{\frac{π}{a}}-\sqrt{\frac{π}{b}}\right)
\end{alignat}

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